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Structure of Atom

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Chapter Notes

Overview
This chapter traces the historical development of atomic models from Dalton's indivisible atom to the quantum mechanical model. It covers the discovery of sub-atomic particles (electron, proton, neutron), the failure of classical physics to explain atomic stability and spectra, and the birth of quantum theory through Planck, Einstein, and Bohr. The one big idea a student must walk away with is that matter and energy are quantized — electrons exist in discrete energy levels described by quantum numbers, and their behaviour is governed by wave-particle duality and the uncertainty principle.
Key Concepts (23)
Cathode rays
Streams of negatively charged particles (electrons) that travel from the cathode to the anode in a cathode ray discharge tube at low pressure and high voltage.
e.g. The glowing beam in a television picture tube is a stream of cathode rays hitting a fluorescent screen.
Electron
A negatively charged sub-atomic particle with absolute charge –1.602176 × 10⁻¹⁹ C and mass 9.109382 × 10⁻³¹ kg (0.00054 u).
e.g. Electrons are the basic constituent of all atoms, as cathode ray properties are independent of the electrode material or gas.
Proton
A positively charged sub-atomic particle found in the nucleus, with absolute charge +1.602176 × 10⁻¹⁹ C and mass 1.6726216 × 10⁻²⁷ kg (1.00727 u).
e.g. The smallest positive ion obtained from hydrogen gas in a discharge tube is called a proton.
Neutron
An electrically neutral sub-atomic particle in the nucleus, with mass 1.674927 × 10⁻²⁷ kg (1.00867 u), discovered by Chadwick in 1932 by bombarding beryllium with α-particles.
e.g. Neutrons have no charge, so they are not deflected by electric or magnetic fields.
Atomic number (Z)
The number of protons in the nucleus of an atom; for a neutral atom, it equals the number of electrons.
e.g. Hydrogen has Z = 1, sodium has Z = 11.
Mass number (A)
The total number of nucleons (protons + neutrons) in the nucleus of an atom.
e.g. For carbon-12, A = 12 (6 protons + 6 neutrons).
Isotopes
Atoms of the same element (same atomic number) having different mass numbers due to different numbers of neutrons.
e.g. Protium (¹H), deuterium (²H), and tritium (³H) are isotopes of hydrogen.
Isobars
Atoms of different elements (different atomic numbers) having the same mass number.
e.g. ¹⁴C and ¹⁴N are isobars.
Thomson model of atom
A model in which the atom is a sphere of uniform positive charge (radius ~10⁻¹⁰ m) with electrons embedded like plums in a pudding or seeds in a watermelon.
e.g. This model explained overall neutrality but failed to explain later scattering results.
Rutherford's nuclear model
A model where the atom has a tiny, dense, positively charged nucleus (radius ~10⁻¹⁵ m) surrounded by electrons moving in circular orbits, held by electrostatic attraction.
e.g. Most α-particles passed through gold foil undeflected, proving most of the atom is empty space.
Electromagnetic radiation
Oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation, travelling at speed c = 3.0 × 10⁸ m/s in vacuum.
e.g. Visible light, radio waves, X-rays are all electromagnetic radiations differing in wavelength and frequency.
Wavelength (λ)
The distance between two successive crests or troughs of a wave; SI unit is metre (m).
e.g. Visible light has wavelengths from 400 nm (violet) to 750 nm (red).
Frequency (ν)
The number of waves that pass a given point in one second; SI unit is hertz (Hz, s⁻¹).
e.g. Vividh Bharati broadcasts at 1368 kHz = 1.368 × 10⁶ Hz.
Wavenumber (ν̄)
The number of wavelengths per unit length; SI unit is m⁻¹, commonly cm⁻¹.
e.g. For yellow light of 5800 Å, wavenumber = 1.724 × 10⁴ cm⁻¹.
Planck's quantum theory
Energy is emitted or absorbed in discrete packets called quanta; energy of a quantum is E = hν, where h = 6.626 × 10⁻³⁴ J s.
e.g. A black body radiates energy only in multiples of hν, not continuously.
Photoelectric effect
The ejection of electrons from a metal surface when light of frequency greater than a threshold frequency (ν₀) strikes it.
e.g. Potassium metal requires ν₀ = 5.0 × 10¹⁴ Hz; red light fails but yellow light succeeds.
Work function (W₀)
The minimum energy required to eject an electron from a metal surface; W₀ = hν₀.
e.g. For sodium, W₀ = 2.3 eV; for copper, W₀ = 4.8 eV.
Photon
A particle of electromagnetic radiation carrying energy E = hν.
e.g. A 400 nm photon has energy 4.969 × 10⁻¹⁹ J.
Line emission spectrum
A spectrum consisting of bright lines at specific wavelengths against a dark background, produced by excited atoms emitting radiation at discrete energies.
e.g. Each element has a unique line spectrum, like a fingerprint.
Continuous spectrum
A spectrum containing all wavelengths in a given range, with no gaps; e.g., white light passed through a prism.
e.g. Rainbow colours from violet to red merging into each other.
Absorption spectrum
A continuous spectrum with dark lines at specific wavelengths where radiation has been absorbed by a sample.
e.g. The dark lines in the solar spectrum correspond to absorption by elements in the Sun's atmosphere.
Black body
An ideal body that absorbs and emits radiation of all frequencies uniformly; approximated by a cavity with a tiny hole.
e.g. Carbon black is a close approximation to a black body.
Dual behaviour of electromagnetic radiation
Light exhibits both wave-like properties (interference, diffraction) and particle-like properties (photoelectric effect).
e.g. In propagation, light behaves as a wave; in interaction with matter, it behaves as a stream of photons.
Formulas & Equations (11)
e/mₑ = 1.758820 × 10¹¹ C kg⁻¹
Reads as: e over m sub e equals 1.758820 times ten to the power 11 coulomb per kilogram
Variables: e = charge on electron (C), mₑ = mass of electron (kg)
Use when: Calculating charge-to-mass ratio of electron; Thomson's experiment.
⚠ Forgetting that this is the ratio, not the individual values of e or mₑ.
e = –1.602176 × 10⁻¹⁹ C
Reads as: e equals negative 1.602176 times ten to the power negative 19 coulomb
Variables: e = charge on electron (C)
Use when: Using Millikan's oil drop result; calculating forces on charged particles.
⚠ Writing the charge as positive; electron charge is negative.
mₑ = 9.109382 × 10⁻³¹ kg
Reads as: m sub e equals 9.109382 times ten to the power negative 31 kilogram
Variables: mₑ = mass of electron (kg)
Use when: Calculating electron mass from e/mₑ and e.
⚠ Confusing mass of electron with mass of proton or neutron.
c = νλ
Reads as: c equals nu lambda
Variables: c = speed of light in vacuum (3.0 × 10⁸ m s⁻¹), ν = frequency (Hz), λ = wavelength (m)
Use when: Converting between frequency and wavelength of electromagnetic radiation.
⚠ Using wrong units; wavelength must be in metres, frequency in Hz.
ν̄ = 1/λ
Reads as: nu bar equals one over lambda
Variables: ν̄ = wavenumber (m⁻¹ or cm⁻¹), λ = wavelength (m or cm)
Use when: Calculating wavenumber from wavelength, especially in spectroscopy problems.
⚠ Forgetting to convert wavelength to cm when using cm⁻¹.
E = hν
Reads as: E equals h nu
Variables: E = energy of a quantum (J), h = Planck's constant (6.626 × 10⁻³⁴ J s), ν = frequency (Hz)
Use when: Calculating energy of a photon or quantum of radiation.
⚠ Using wavelength instead of frequency; must convert using c = νλ.
E = hc/λ
Reads as: E equals h c over lambda
Variables: E = energy of photon (J), h = Planck's constant (6.626 × 10⁻³⁴ J s), c = speed of light (3.0 × 10⁸ m s⁻¹), λ = wavelength (m)
Use when: Calculating photon energy directly from wavelength.
⚠ Forgetting to convert nm to m; 1 nm = 10⁻⁹ m.
½ mₑ v² = hν – hν₀
Reads as: half m sub e v squared equals h nu minus h nu zero
Variables: mₑ = mass of electron (kg), v = velocity of ejected electron (m s⁻¹), h = Planck's constant (J s), ν = frequency of incident light (Hz), ν₀ = threshold frequency (Hz)
Use when: Calculating kinetic energy of photoelectron in photoelectric effect.
⚠ Forgetting that hν₀ is the work function; writing kinetic energy = hν only.
Atomic number (Z) = number of protons = number of electrons in neutral atom
Reads as: Atomic number Z equals number of protons equals number of electrons in a neutral atom
Variables: Z = atomic number (dimensionless)
Use when: Determining number of protons or electrons from atomic number.
⚠ Assuming Z equals number of neutrons; it does not.
Mass number (A) = number of protons (Z) + number of neutrons (n)
Reads as: Mass number A equals number of protons Z plus number of neutrons n
Variables: A = mass number, Z = atomic number, n = number of neutrons
Use when: Calculating number of neutrons from mass number and atomic number.
⚠ Confusing mass number with atomic mass; mass number is an integer.
q = n e
Reads as: q equals n e
Variables: q = charge on oil droplet (C), n = integer (1, 2, 3...), e = elementary charge (1.602 × 10⁻¹⁹ C)
Use when: Millikan's oil drop experiment; charge is quantized.
⚠ Forgetting that n is an integer; charge is always an integral multiple of e.
How It Works
Cathode ray discharge tube experiment
  1. A glass tube with two metal electrodes is partially evacuated to low pressure.
  2. A high voltage is applied across the electrodes.
  3. Current flows as a stream of particles from cathode to anode — these are cathode rays.
  4. Cathode rays travel in straight lines, are deflected by electric and magnetic fields, and cause fluorescence on a zinc sulphide screen.
  5. The properties of cathode rays are independent of the electrode material and the gas in the tube.
💡 Cathode rays are electrons; they are negatively charged and are emitted from the cathode.
Thomson's determination of e/mₑ
  1. A cathode ray tube is set up with both electric and magnetic fields perpendicular to each other and to the electron path.
  2. With only electric field, electrons deflect to point A on the screen.
  3. With only magnetic field, electrons deflect to point C.
  4. The fields are balanced so electrons hit the centre point B (no net deflection).
  5. From the field strengths and deflection, the charge-to-mass ratio e/mₑ is calculated.
💡 The deflection depends on charge magnitude, mass of particle, and field strength.
Millikan's oil drop experiment
  1. Oil droplets are sprayed as mist through a tiny hole into an electrical condenser.
  2. The downward motion of droplets is viewed through a telescope with a micrometer eyepiece.
  3. The air is ionized by X-rays, giving droplets a charge through collision with ions.
  4. By adjusting the voltage, the droplet's motion can be retarded, accelerated, or made stationary.
  5. The charge on each droplet is always an integral multiple of e (q = ne).
💡 Charge is quantized; the smallest unit is the charge on one electron.
Rutherford's α-particle scattering experiment
  1. A beam of high-energy α-particles from a radioactive source is directed at a thin gold foil (thickness ~100 nm).
  2. A circular fluorescent zinc sulphide screen around the foil detects flashes when α-particles strike it.
  3. Most α-particles pass through the foil undeflected.
  4. A small fraction are deflected by small angles.
  5. A very few (about 1 in 20,000) bounce back, deflected by nearly 180°.
💡 Most of the atom is empty space; the positive charge is concentrated in a tiny nucleus.
Photoelectric effect (Einstein's explanation)
  1. Light of frequency ν strikes a metal surface; each photon carries energy hν.
  2. If ν ≥ ν₀ (threshold frequency), an electron absorbs the photon's energy instantaneously.
  3. The electron uses energy hν₀ (work function) to escape the metal.
  4. The remaining energy (hν – hν₀) becomes the kinetic energy of the ejected electron: ½ mₑ v² = hν – hν₀.
  5. Increasing light intensity increases number of photons, hence number of electrons, but not their kinetic energy.
💡 Kinetic energy of photoelectrons depends on frequency, not intensity; intensity affects only the number of electrons.
Production of emission spectrum
  1. Energy is supplied to a sample (atom, molecule, or ion) by heating or irradiating it — the sample becomes 'excited'.
  2. Excited atoms are unstable and return to lower energy states by emitting radiation.
  3. The emitted radiation is passed through a prism or diffraction grating.
  4. Only specific wavelengths (bright lines) are observed, corresponding to energy differences between quantized levels.
  5. Each element has a unique line emission spectrum.
💡 Line spectra prove that electronic energy levels are quantized — not all energies are allowed.
Production of absorption spectrum
  1. A continuous spectrum of radiation (all wavelengths) is passed through a sample.
  2. The sample absorbs radiation at specific wavelengths corresponding to energy transitions.
  3. The transmitted radiation is analysed with a spectroscope.
  4. Dark lines appear in the continuous spectrum at the wavelengths that were absorbed.
  5. The absorption spectrum is like the photographic negative of the emission spectrum.
💡 Absorption lines occur at the same wavelengths as emission lines for a given element.
Exam Traps — Don't Make These Mistakes
⚠ Confusing cathode rays with canal rays
✗ Cathode rays are positively charged particles moving from anode to cathode.
→ Cathode rays are negatively charged electrons moving from cathode to anode. Canal rays (positive rays) move from anode to cathode and their mass depends on the gas.
✓ Cathode rays are streams of electrons (negatively charged) that travel from the cathode to the anode in a discharge tube.
⚠ Writing the charge on electron as positive
✗ The charge on an electron is +1.6 × 10⁻¹⁹ C.
→ The electron is negatively charged. The proton carries the positive charge of the same magnitude.
✓ The charge on an electron is –1.602176 × 10⁻¹⁹ C.
⚠ Mixing up atomic number and mass number in notation
✗ For ³⁵₁₇Cl, the atomic number is 35 and mass number is 17.
→ In the notation ᴬ₂X, the superscript A is the mass number and the subscript Z is the atomic number. So for ³⁵₁₇Cl, A = 35, Z = 17.
✓ For ³⁵₁₇Cl, the atomic number (Z) is 17 and the mass number (A) is 35.
⚠ Thinking photoelectric effect depends on light intensity for kinetic energy
✗ Brighter light ejects electrons with higher kinetic energy.
→ Kinetic energy of photoelectrons depends on the frequency (ν) of light, not its intensity. Intensity determines the number of photons and hence the number of electrons ejected.
✓ The kinetic energy of ejected electrons is given by ½ mₑ v² = h(ν – ν₀); it increases with frequency, not intensity.
⚠ Confusing wavenumber with frequency
✗ Wavenumber is the number of waves per second.
→ Wavenumber (ν̄) is the number of wavelengths per unit length (1/λ), not per second. Frequency (ν) is the number of waves per second.
✓ Wavenumber is defined as the number of wavelengths per unit length, i.e., ν̄ = 1/λ, with units m⁻¹ or cm⁻¹.
⚠ Forgetting that neutrons have no charge but have mass
✗ Neutrons have no charge and negligible mass.
→ Neutrons are electrically neutral but their mass (1.00867 u) is slightly greater than that of a proton (1.00727 u). They contribute significantly to the mass of the nucleus.
✓ Neutrons have no charge (0 C) and a mass of 1.674927 × 10⁻²⁷ kg (1.00867 u), slightly more than a proton.
Compare & Contrast
Distribution of positive charge
Thomson model of atom
Rutherford's nuclear model
Thomson proposed positive charge is uniformly spread over the entire atom (sphere of radius ~10⁻¹⁰ m), while Rutherford showed it is concentrated in a tiny nucleus (radius ~10⁻¹⁵ m).
How the spectrum is produced
Emission spectrum
Absorption spectrum
Emission spectrum is produced by excited atoms emitting radiation at specific wavelengths (bright lines on dark background). Absorption spectrum is produced when a continuous spectrum passes through a sample and specific wavelengths are absorbed (dark lines on bright background).
Atomic number and mass number
Isotopes
Isobars
Isotopes have same atomic number (Z) but different mass numbers (A) — e.g., ¹²C and ¹⁴C. Isobars have same mass number (A) but different atomic numbers (Z) — e.g., ¹⁴C and ¹⁴N.
Experimental evidence
Wave nature of light
Particle nature of light
Wave nature explains interference and diffraction. Particle nature explains black body radiation and photoelectric effect. Light has dual behaviour — both wave and particle properties.
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1 short answer (2-3 marks) on photoelectric effect or Rutherford model, 1 long answer (5 marks) on atomic spectra or quantum theory. Total 8-10 marks.
Most asked: Rutherford's α-particle scattering experiment and conclusions · Photoelectric effect and Einstein's equation · Planck's quantum theory and E = hν · Bohr's model (though not fully covered in this text, it is a major theme) · Atomic number, mass number, isotopes, isobars · Electromagnetic spectrum and c = νλ · Dual behaviour of electromagnetic radiation
Don't skip: Derive or explain the photoelectric effect with Einstein's equation (½ mₑ v² = hν – hν₀).; Describe Rutherford's α-particle scattering experiment, observations, and conclusions.; Numerical problems using c = νλ, E = hν, and E = hc/λ.; Explain Planck's quantum theory and its application to black body radiation.
One-Line Revision (read aloud before the exam)
  1. Cathode rays are streams of electrons that travel from cathode to anode in a discharge tube.
  2. J.J. Thomson measured the charge-to-mass ratio (e/mₑ) of the electron as 1.758820 × 10¹¹ C kg⁻¹.
  3. Millikan's oil drop experiment proved that charge is quantized: q = n e, with e = 1.602 × 10⁻¹⁹ C.
  4. Protons are positively charged particles in the nucleus; neutrons are neutral and were discovered by Chadwick.
  5. Rutherford's α-particle scattering experiment showed that most of the atom is empty space with a tiny, dense, positively charged nucleus.
  6. Atomic number (Z) = number of protons = number of electrons in a neutral atom.
  7. Mass number (A) = number of protons + number of neutrons.
  8. Isotopes have same Z but different A; isobars have same A but different Z.
  9. Electromagnetic radiation travels at c = 3.0 × 10⁸ m/s in vacuum and obeys c = νλ.
  10. Wavenumber ν̄ = 1/λ, commonly expressed in cm⁻¹.
  11. Planck's quantum theory: energy is emitted or absorbed in discrete quanta, E = hν, where h = 6.626 × 10⁻³⁴ J s.
  12. Photoelectric effect: electrons are ejected from a metal when light of frequency ν ≥ ν₀ strikes it; kinetic energy = h(ν – ν₀).
  13. Work function W₀ = hν₀ is the minimum energy needed to eject an electron.
  14. Light has dual behaviour: wave nature (interference, diffraction) and particle nature (photoelectric effect).
  15. Line emission spectra consist of bright lines at specific wavelengths, proving quantized energy levels.
  16. Absorption spectra show dark lines on a continuous background at wavelengths absorbed by the sample.
  17. A black body is an ideal absorber and emitter of radiation; its radiation spectrum depends only on temperature.
  18. The Thomson model (plum pudding) failed to explain α-particle scattering; Rutherford's model explained it but could not account for atomic stability.
  19. The charge on an electron is –1.602176 × 10⁻¹⁹ C and its mass is 9.109382 × 10⁻³¹ kg.
  20. The mass of a proton is 1.6726216 × 10⁻²⁷ kg and of a neutron is 1.674927 × 10⁻²⁷ kg.

MCQs — 10 Questions with Answers

Q1. Which of the following statements about cathode rays is NOT correct?

  • A. They travel in straight lines in the absence of electric or magnetic field
  • B. They are deflected towards the positive electrode in an electric field
  • C. Their properties depend on the material of the cathode used ✓
  • D. They cause fluorescence when hitting zinc sulphide coating

Answer: C — Cathode ray properties are independent of electrode material and gas type, proving electrons are universal constituents of all atoms.

Q2. The charge-to-mass ratio (e/me) of an electron is 1.758820 × 10¹¹ C kg⁻¹. If the charge on electron is –1.602176 × 10⁻¹⁹ C, what is the mass of electron?

  • A. 9.11 × 10⁻³¹ kg ✓
  • B. 1.76 × 10⁻¹² kg
  • C. 2.82 × 10⁻⁵⁰ kg
  • D. 5.64 × 10⁻¹¹ kg

Answer: A — me = e ÷ (e/me) = 1.602176 × 10⁻¹⁹ ÷ 1.758820 × 10¹¹ = 9.11 × 10⁻³¹ kg.

Q3. In Thomson's e/me measurement apparatus, the electrons hit point B on the screen when both electric and magnetic fields are applied. What does this indicate?

  • A. The electric force is greater than the magnetic force
  • B. The electric force is less than the magnetic force
  • C. The electric force and magnetic force are balanced ✓
  • D. The electron has no charge

Answer: C — Point B is on the original path (no field), so electrons experience no net force; thus qE = qvB (forces are balanced).

Q4. Millikan's oil drop experiment directly measures which property of electron?

  • A. Charge on electron ✓
  • B. Mass of electron
  • C. Charge-to-mass ratio
  • D. Spin of electron

Answer: A — The oil drop experiment balances gravitational and electric forces on a suspended charged oil drop, allowing calculation of the electron's charge.

Q5. If the deflection of electrons in an electric field increases when the field strength increases, and decreases when electron mass increases, which statement best explains this observation?

  • A. Deflection is inversely proportional to both field strength and electron mass
  • B. Deflection is directly proportional to field strength and inversely proportional to electron mass ✓
  • C. Deflection is directly proportional to both field strength and electron mass
  • D. Deflection depends only on field strength, not on electron mass

Answer: B — From F = qE and a = F/m, deflection ∝ E/m, so deflection increases with field strength and decreases with mass.

Q6. A student observes that when a cathode ray tube is evacuated to different pressures, the properties of the rays remain unchanged. This observation proves that:

  • A. Electrons can only exist in vacuum
  • B. Electrons are fundamental constituents of all atoms ✓
  • C. The anode repels electrons more strongly at lower pressures
  • D. Electron mass varies with pressure

Answer: B — Since cathode ray properties are independent of gas type and pressure, electrons must be universal building blocks of all matter.

Q7. In Faraday's electrolysis experiments, chemical reactions occurred at the electrodes. This observation was significant because it suggested:

  • A. Electricity is continuous like water flow
  • B. Electricity has particulate nature ✓
  • C. All chemicals are electrical conductors
  • D. Electrons are negatively charged particles

Answer: B — Discrete chemical reactions at specific electrodes suggested electricity consists of discrete particles, not continuous waves.

Q8. Consider two electrons: one deflected by 2 cm and another by 4 cm in the same electric field. Assuming both have the same charge, which statement is true?

  • A. Both electrons have the same mass
  • B. The first electron has twice the mass of the second ✓
  • C. The second electron has twice the mass of the first
  • D. The mass difference cannot be determined from deflection data alone

Answer: B — Greater deflection (4 cm) indicates smaller mass; since deflection ∝ 1/m, the electron deflected 2 cm has twice the mass of the one deflected 4 cm.

Q9. Both Thomson's e/me measurement and Millikan's charge measurement are important. Which statement correctly relates these two experiments?

  • A. Thomson measured charge; Millikan measured e/me ratio
  • B. Both measured the same quantity using different methods
  • C. Thomson measured e/me; Millikan measured charge, allowing calculation of electron mass ✓
  • D. Millikan's experiment was unnecessary because Thomson already found the electron mass

Answer: C — Thomson found e/me = 1.76 × 10¹¹ C/kg; Millikan found e = 1.6 × 10⁻¹⁹ C; combining them gives me = 9.11 × 10⁻³¹ kg.

Q10. A cathode ray tube experiment shows that when voltage across electrodes increases, electron deflection in a perpendicular electric field also increases. This is because: (A) Electrons gain higher velocity, so they spend less time in the field. (B) Higher voltage creates stronger deflecting field, so electrons experience greater force. (C) Electrons become heavier at higher voltages. (D) The relationship between voltage and deflection is inversely proportional.

  • A. Electrons gain higher velocity, so they spend less time in the field
  • B. Higher voltage creates stronger deflecting field, so electrons experience greater force ✓
  • C. Electrons become heavier at higher voltages
  • D. The relationship between voltage and deflection is inversely proportional

Answer: B — Higher voltage between electrodes increases the electric field strength, which increases the force on electrons (F = qE), causing greater deflection.

Flashcards

What is the charge-to-mass ratio (e/me) of an electron determined by Thomson?

1.758820 × 10¹¹ C/kg, measured using crossed electric and magnetic fields in a cathode ray tube.

What is the magnitude of the charge on a single electron?

1.602176 × 10⁻¹⁹ C (or approximately 1.6 × 10⁻¹⁹ C), determined by Millikan's oil drop experiment.

What is the mass of an electron in kilograms?

9.1094 × 10⁻³¹ kg, calculated from e/me ratio and the known electron charge.

In a cathode ray tube, what do cathode rays consist of?

Negatively charged particles called electrons, which flow from the cathode (negative electrode) to the anode (positive electrode).

What is the key observation from Faraday's electrolysis experiments?

Chemical reactions at electrodes when electricity passes through electrolyte solutions, suggesting the particulate nature of electricity.

How did Thomson balance the deflection of electrons in crossed electric and magnetic fields?

By adjusting field strengths until electrons travelled straight without deflection, allowing calculation of the e/me ratio.

Why do cathode ray properties not depend on electrode material or gas type?

Because electrons are fundamental constituents of all atoms, so they are identical regardless of the substance used.

What does Millikan's oil drop experiment measure and how?

Measures the charge on an electron by balancing gravitational and electric forces on a suspended charged oil drop.

What happens to an electron's deflection in an electric field if its mass decreases?

The deflection increases because a lighter particle experiences greater acceleration from the same electric force.

Which scientist first showed that electricity has particulate nature through electrolysis?

Michael Faraday in 1830, who observed chemical reactions at electrodes and formulated laws of electrolysis.

Important Board Questions

Define the term 'electron' and state any two of its characteristic properties discovered through cathode ray experiments. [2 marks]

Electron is a negatively charged subatomic particle. State two from: (1) travels from cathode to anode, (2) properties independent of electrode/gas material, (3) deflected in electric/magnetic fields, (4) charge = –1.6 × 10⁻¹⁹ C.

Explain how Thomson determined the charge-to-mass ratio (e/me) of electron using a cathode ray tube with crossed electric and magnetic fields. Include a description of the key principle used to balance the forces. [5 marks]

Describe: (1) setup with perpendicular E and B fields, (2) deflection at point A (E-field only) and point C (B-field only), (3) balancing condition: qE = qvB at point B (straight path), (4) calculation of e/me from field strengths and electron velocity. Show working: e/me ∝ E/B².

Millikan's oil drop experiment is a landmark achievement in determining fundamental properties of matter. Explain the principle of the experiment, derive the condition for a stationary oil drop, and calculate the number of electrons on a drop if the charge on the drop is –3.2 × 10⁻¹⁹ C (given: charge on one electron = –1.6 × 10⁻¹⁹ C). [6 marks]

Explain: (1) setup with charged oil drop suspended between parallel metal plates, (2) forces acting: weight (mg) downward and electric force (qE) upward, (3) equilibrium condition: mg = qE, (4) from this, charge q can be found, (5) number of electrons = total charge ÷ charge per electron. Calculate: n = 3.2 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 2 electrons. Connect to why this proved quantization of charge.

Assertion-Reason Questions for Structure of Atom

Q1. Assertion (A):

Cathode rays are deflected toward the positive electrode in an electric field.

Reason (R):

Cathode rays consist of positively charged particles that are attracted to negative electrodes.

Answer: C — A is true but R is false — The assertion is true—cathode rays do deflect toward the positive electrode. However, the reason is false because cathode rays consist of negatively charged electrons, which are attracted to positive electrodes, not toward negative ones. The reason reverses the charge polarity.

Q2. Assertion (A):

The charge-to-mass ratio (e/me) of electrons determined by Thomson is independent of the gas used in the cathode ray tube.

Reason (R):

Electrons are fundamental particles present in all atoms, and their intrinsic properties do not depend on the surrounding gas environment.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the e/me ratio is invariant. The universality of the e/me value across different gases proved that electrons are fundamental constituents of all matter, not gas-dependent artifacts.

Q3. Assertion (A):

In Millikan's oil drop experiment, the charge on an electron was determined to be –1.602176 × 10⁻¹⁹ C.

Reason (R):

This measurement was made by balancing the electric force on charged oil droplets against their gravitational force.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—Millikan did measure the electron charge with this value. The reason is also true—he did balance electric force against gravity. However, the reason does not fully explain why this determines the electron charge; it omits that quantization of charge on droplets revealed the elementary charge unit. Both are true but the causal link is incomplete.

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Case-Based Questions (CBQ) for Structure of Atom

Case Study:

A research team in Bangalore studied the behavior of cathode rays in a modernized discharge tube setup. They applied a variable voltage between 1000 V and 5000 V across electrodes separated by 5 cm in a tube filled with argon gas at 0.01 mmHg pressure. When 2000 V was applied, electrons reached the anode creating a steady glow. When voltage increased to 4000 V, the glow intensity increased significantly and the electrons struck the anode with nearly double the kinetic energy. However, when pressure was increased to 0.05 mmHg at the same 4000 V, the glow diminished noticeably despite the higher voltage. The scientists attributed this to increased collision frequency between electrons and argon atoms.

Q1. Why did the glow intensity increase when voltage doubled from 2000 V to 4000 V?

  • A. Electrons gained twice the kinetic energy and more electrons reached the anode per unit time
  • B. The electric field strength increased, accelerating electrons to higher speeds and increasing collision-induced fluorescence at the anode ✓
  • C. Argon atoms became ionized at higher voltages
  • D. The electrode separation decreased with higher voltage

Answer: B

Q2. The kinetic energy of electrons at the anode is approximately proportional to the applied voltage. If KE at 2000 V was E₀, what is the approximate KE at 4000 V?

  • A. E₀
  • B. 2E₀ ✓
  • C. 4E₀
  • D. E₀/2

Answer: B

Q3. When pressure increased from 0.01 mmHg to 0.05 mmHg at constant 4000 V, the glow diminished. This occurred because:

  • A. More gas molecules increased collision probability, reducing the number of electrons reaching the anode without losing significant energy ✓
  • B. Higher pressure increased the voltage required to accelerate electrons
  • C. Higher pressure caused electrons to reverse direction toward the cathode
  • D. The argon gas itself emitted light, blocking the visible glow from the anode

Answer: A

Q4. To restore the glow intensity at 0.05 mmHg pressure to match that observed at 0.01 mmHg and 4000 V, the team would most likely need to:

  • A. Increase the voltage further to overcome collision losses and ensure sufficient electrons reach the anode ✓
  • B. Decrease the voltage to reduce electron speed
  • C. Replace argon with helium gas
  • D. Reduce the electrode separation

Answer: A

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